Grass Bush Grass Bush Grass Bush

1
Black Green Orange Red Yellow

5 9 4 1 1
2
Black Green Orange Red

3 8 6 3
3
Black Green Orange Yellow

8 9 2 1
4
Black Green Orange Red Yellow

4 5 3 3 5
5
Black Green Orange Yellow

8 6 4 2
6
Black Green Orange Yellow

6 8 4 2

Population
Black
Green
Orange
Red
Yellow
Total
1
Grass
5
9
4
1
1
20
2
Bush
3
8
6
3
0
20
3
Grass
8
9
2
0
1
20
4
Bush
4
5
3
3
5
20
5
Grass
8
6
4
0
2
20
6
Bush
6
8
4
0
2
20

Total
34
45
23
7
11
120


Black
Green
Orange
Red
Yellow
Expected value (E)
5.666
7.5
3.833
1.166
1.833

  • Due to our knowledge of jelly babies we have decided to combine red with black and green with yellow because we feel they would be predated on very similarly. Also our expected values for red and yellow are very low therefore need to be combined.

  • The null hypothesis is that there will be no significant difference in the population frequencies between the different habitats.
  • We decided to compare the proportion of polymorphisms across our samples.
  • At a p value of 0.05, our null hypothesis will be true if we obtain a chi squared value which is greater than 0.05. However if the p value is equal to or less than to 0.05, it will suggest that our results will be significant.

Population
Black + Red
Green + Yellow
Orange
Total
1
Grass
6
10
4
20
2
Bush
6
8
6
20
3
Grass
8
10
2
20
4
Bush
7
10
3
20
5
Grass
8
8
4
20
6
Bush
6
10
4
20

Total
41
56
23
120


Black + Red
Green + Yellow
Orange
Expected Value (E)
6.833
9.333
3.833

  • Now our expected values are significant enough for Chi square test

Chi Square Test

All six populations:

Population
Black + Red
Green + Yellow
Orange
Total
X2 1
Grass
0.102
0.048
0.007
0.157
X2 2
Bush
0.102
0.190
1.225
1.517
X2 3
Grass
0.199
0.048
0.877
1.124
X2 4
Bush
0.004
0.048
0.181
0.233
X2 5
Grass
0.199
0.190
0.007
0.396
X2 6
Bush
0.102
0.048
0.007
0.157

Total
0.708
0.572
2.304
3.584

Grass:

Population
Black + Red
Green + Yellow
Orange
Total
1
Grass
6
10
4
20
3
Grass
8
10
2
20
5
Grass
8
8
4
20

Total
22
28
10



Black + Red
Green + Yellow
Orange
Expected Value (E)
7.333
9.333
3.333


Population
Black + Red
Green + Yellow
Orange
Total
X2 1
Grass
0.242
0.048
0.133
0.423
X2 3
Grass
0.061
0.048
0.533
0.642
X2 5
Grass
0.061
0.190
0.133
0.384

Total
0.364
0.286
0.799
1.449

DF
4
Critical Value (0.05)
9.488
Chi Squared value
1.449
  • The chi squared value of 1.449, is significantly lower than the critical value of 9.488 therefore we can accept the null hypothesis which is that the difference in allele frequencies is not significant between populations. therefore the polymorphism of snails may be attributed to selection.

Bush:

Population
Black + Red
Green + Yellow
Orange
Total
2
Bush
6
8
6
20
4
Bush
7
10
3
20
6
Bush
6
10
4
20

Total
19
28
13



Black + Red
Green + Yellow
Orange
Expected Value (E)
6.333
9.333
4.333


Population
Black + Red
Green + Yellow
Orange
Total
2
Bush
0.018
0.190
0.641
0.849
4
Bush
0.070
0.048
0.410
0.528
6
Bush
0.018
0.048
0.026
0.092

Total
0.106
0.286
1.077
1.469

DF
4
Critical Value (0.05)
9.488
Chi Squared value
1.469

  • Our chi squared value of 1.469 is significantly lower than our critical value of 9.488 and therefore we can conclude that our null hypothesis can be accepted. Our null hypothesis suggests that the allele frequencies are not significant between populations.